# The sound of Decibels

1. Nov 29, 2004

### mikej_45

Ok so will someone please explain to me why it is that when you have a machine that produces 70 decibels in a room and add an identical machine that also produces 70 decibels to that room, the decibels don't double... my book is very vague about this and does not explain it well... any help would be appreciated. Thank you!!

2. Nov 30, 2004

### Gokul43201

Staff Emeritus
The decibel is a unit proportional to the logarithm of the sound power level. When you have two sources of sound, the total power is typically the sum of the two individual powers, but the logarithm of a sum of numbers is not equal to the sum of the logarithms.

log(a+b) is not the same as log(a) + log(b)

So you can not expect the decibel level to be the sum of the individual decibel levels. But you can calculate the total decibel level from the individual ones.

In this case, it will be 70 + 10log(2). Do you understand why ?

Last edited: Nov 30, 2004
3. Nov 30, 2004

### ricemike

explain logarithm in laymans terms please

4. Nov 30, 2004

### mikej_45

Umm...thats about what my book says...I do know that the decibels won't double...I guess I just can't figure out how to do the math to get it to work.

So If the inensity level of one machine is 70 dB(decibels)

β = 10 log (1.0×10-5 W/m2) = 70 dB (decibels)
1.0×10-12 W/m2

(-5) - (-12) = 7 so you get 10 to the power of 7...70 dB

and when you add one more machine it doubles the energy input into the sound...thereby doubling the intensity:

my problem is that I don't understand how this works:

β = 10 log (2.0×10-5 W/m2) = 73 dB (decibels)
1.0×10-12 W/m2

How do these numbers come anywhere close to 73 dB? Maybe I just don't understand logs...anyone able to explain this in simplified terms...

5. Nov 30, 2004

### Gokul43201

Staff Emeritus
$$\beta = 10~log( \frac{2.0*10^{-5}} {1.0*10^{-12}}) = 10~log(2.0*10^7) = 10*[log(2) + log(10^7)] = 10log(2) + 70 = 3.01 + 70 = 73.01~dB$$

Remember the rules for logarithms : log(a*b) = log(a) + log(b) and log(a/b) = log(a) - log(b)

Last edited: Nov 30, 2004
6. Dec 1, 2004

### mikej_45

Thanks for the help, sometimes its as easy as seeing it done I guess. Thanks again.

7. Dec 3, 2004

### wire2

Simply put, doubling the energy level (audio or radio frequency) adds 3 Db. So doubling 3 Db = 6 Db, doubling 70 Db = 73 Db. The same rules apply to calculating the gain of directional antennas.

8. Dec 6, 2004

### Cliff_J

Logs are used because its easy to describe the ratio and its represents how our body responds to the sound.

0 db is defined as the lowest threshold of hearing, its at about the level of pressure the air molecules bouncing off the eardrum create or just plain incredibly quiet.

All sound is referenced to that level. But rather than say that this machine is 10 million times the reference level and that if you add another machine then its 20 million times reference this still doesn't mean too much. What if the next machine is 100 million times the reference level? How can you compare those huge numbers? What do they mean to our ears?

Instead you say the machine is 70 db and if you add the other its 73 db. And if it was 100 million times larger than reference its 80 db.

The term decibel means one-tenth of a bel. The difference in level between 70db and 80db is one bel and to you that would be about equal to twice as loud. The smallest level difference noticeable is one decibel and hence why its used to give more distinction between the doubling/halving of sound levels.

The threshold of pain for sound levels is 120 db which is 10 quadrillion times the reference level of pressure. A 747 plane is 100 times that level of pressure at 140db. Our hearing has a very large range of pressures it can respond to and using decibels makes it easy to understand (hopefully) in terms that can easily be compared.

There's more to it but that should help explain the mystery a little better in layman's terms.

Cliff